Question

The monthly saving of 500 students is normally distributed with mean RM200 and standard deviation RM150.

a) find the number of students has saving between RM180 and RM281 a month.

b) determine the value of X given that 85% of the students saving more than RMx a month.

Answer 1

Given :

Sample size

n = 500

Mean

μ = 200

Standard deviation

σ = 0

a)

P(180 < x < 281)

Using z-score formula ,

$z =\frac{x-μ}{σ}\\ =P(\frac{180 – 200}{150}< \frac{x-μ}{σ}< \frac{281 – 200}{150})\\ = P(-0.13 < z < 0.54)\\ = P (< 0.54) - P(Z <-0.13)$

Using Excel function, =NORMSDIST (z)

=NORMSDIST (0.54), = 0.7054

=NORMSDIST (-0.13), = 0.4483

=0.7054 - 0.4483

=0.2571

P(180 < x < 281) = 0.2571

500*0.2571 = 128.55

Rounding to nearest whole number would be 129

Therefore the number of students has saving between $ 180 and $ 281 a month are 129

That is 129 students has saving between $ 180 and $ 281 a month.

b)

P(X>x)=0.85

P(X<x) =1-0.85

P(X<x)=0.15

Using Excel function, =NORMSINV (area)

=NORMSINV (0.15)

=-1.04 (rounded to two decimals)

z = -1.04

Using the z-score formula ,

$z=\frac{x-μ}{σ}$

x= μ+zσ

x= 200+ (-1.04 * 150)

x= 200+ (-156)

x= 44